Find the equation of the circle circumscribing the triangle whose sides are $\;$ $5x - 3y + 4 = 0$, $\;$ $2x + 3y - 5 = 0$, $\;$ $x - y = 0$
The given equations are
$5x - 3y + 4 = 0$, $\;$ $2x + 3y - 5 = 0$, $\;$ $x - y = 0$
The coordinates of the points of intersection of any two of the three given straight lines satisfy the equation
$\lambda \left(5x - 3y + 4\right) \left(2x + 3y - 5\right) + \mu \left(2x + 3y - 5\right) \left(x - y\right)$
$\hspace{2cm} + \nu \left(x - y\right) \left(5x - 3y + 4\right) = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ represents a locus which passes through the vertices of the triangle found by the given lines.
This locus will be a circle if
the coefficients of $x^2$ and $y^2$ are equal
and the coefficient of $xy$ is zero.
i.e. $\;$ $10 \lambda + 2 \mu + 5 \nu = -9 \lambda - 3 \mu + 3 \nu$
i.e. $\;$ $19 \lambda + 5 \mu + 2 \nu = 0$ $\;\;\; \cdots (2)$
and $\;$ $15 \lambda - 6 \lambda - 2 \mu + 3 \mu - 3 \nu - 5 \nu = 0$
i.e. $\;$ $9 \lambda + \mu - 8 \nu = 0$ $\;\;\; \cdots (3)$
$\therefore \;$ $\dfrac{\lambda}{\left(1^2 + \left(-1\right)^2 \right) \left(2 \times \left(-3\right) - 5 \times 3\right)} = \dfrac{\mu}{\left(5^2 + \left(-3\right)^2\right) \left(1 \times 3 - 2 \times \left(-1\right)\right)}$
$\hspace{3.5cm} = \dfrac{\nu}{\left(2^2 + 3^2\right) \left(5 \times \left(-1\right) - 1 \times \left(-3\right)\right)}$
i.e. $\;$ $\dfrac{\lambda}{-42} = \dfrac{\mu}{170} = \dfrac{\nu}{-26}$
Hence the required equation of circle is
$-42 \left(5x - 3y + 4\right) \left(2x + 3y - 5\right) + 170 \left(2x + 3y - 5\right) \left(x - y\right)$
$\hspace{2.5cm} - 26 \left(x - y\right) \left(5x - 3y + 4\right) = 0$
i.e. $\;$ $-420 x^2 - 630 xy + 1050 x + 252 xy + 378 y^2 - 630y - 336x - 504y + 840$
$\hspace{1cm}$ $340 x^2 - 340 xy + 510 xy - 510 y^2 - 850x + 850y$
$\hspace{1.5cm}$ $-130 x^2 + 78 xy - 104 x + 130 xy - 78y^2 + 104 y = 0$
i.e. $\;$ $-210 x^2 - 210 y^2 - 240x - 180y + 840 = 0$
i.e. $\;$ $-7x^2 - 7y^2 - 8x - 6y + 26 = 0$